Let a, b, x, y be real numbers such that ` a^(2)+b^(2)=81, x^(2)+y^(2)=121 and ax + by = 99` . Then the set of all possible values of ay - bx is -

Options

A. `(0,9/11]`
B. `(0,9/11)`
C. `{0}`
D. `[9/11,infty)`

Explanation

To solve the problem, we need to find the set of all possible values of \( ay - bx \) given the conditions:

1. \( a^2 + b^2 = 81 \)
2. \( x^2 + y^2 = 121 \)
3. \( ax + by = 99 \)

We can use the Cauchy-Schwarz inequality to relate these equations.

### Step 1: Apply Cauchy-Schwarz Inequality

Using the Cauchy-Schwarz inequality, we have:

\[ (ax + by)^2 \leq (a^2 + b^2)(x^2 + y^2) \]

Substituting the known values:

\[ 99^2 \leq (81)(121) \]

Calculating both sides:

\[ 9801 \leq 9801 \]

This shows that the equality holds, which means that the vectors \( (a, b) \) and \( (x, y) \) are proportional.

### Step 2: Find the Proportionality Constant

Since the equality holds, we can write:

\[ \frac{a}{x} = \frac{b}{y} = k \text{ (some constant)} \]

From this, we can express \( a \) and \( b \) in terms of \( x \) and \( y \):

\[ a = kx \quad \text{and} \quad b = ky \]

### Step 3: Substitute into the First Equation

Substituting \( a \) and \( b \) into the equation \( a^2 + b^2 = 81 \):

\[ (kx)^2 + (ky)^2 = 81 \]

This simplifies to:

\[ k^2(x^2 + y^2) = 81 \]

Since \( x^2 + y^2 = 121 \):

\[ k^2 \cdot 121 = 81 \]

Solving for \( k^2 \):

\[ k^2 = \frac{81}{121} = \left(\frac{9}{11}\right)^2 \]

Thus, \( k = \frac{9}{11} \) or \( k = -\frac{9}{11} \).

### Step 4: Express \( ay - bx \)

Now, we can express \( ay - bx \):

\[ ay - bx = (kx)y - (ky)x = k(yy - xx) = k(y^2 - x^2) \]

### Step 5: Calculate \( y^2 - x^2 \)

Using the values \( x^2 = 121 \) and \( y^2 = 121 \):

\[ y^2 - x^2 = 121 - 121 = 0 \]

Thus, we have:

\[ ay - bx = k \cdot 0 = 0 \]

### Conclusion

The only possible value for \( ay - bx \) is:

\[ \boxed{0} \]